<span>Answer:
Chemical equations are balanced in order to: 1) satisfy the Law of Conservation of Mass, and 2) establish the mole relationships needed for stoichiometric calculations. The Law of Conservation of Mass: The Law of Conservation of Mass states that mass cannot be created or destroyed.</span>
Answer:
3.49 g
Explanation:
The mass is the product of volume and density:
(8.96 g/cm³)(0.39 cm³) ≈ 3.49 g
The mass of a pure-copper penny would be 3.49 g.
Answer:
9.35g
Explanation:
The molarity equation establishes that:
![\textrm{molarity}=\frac{\textrm{moles of solute}}{\textrm{liters of solution}}](https://tex.z-dn.net/?f=%5Ctextrm%7Bmolarity%7D%3D%5Cfrac%7B%5Ctextrm%7Bmoles%20%20of%20solute%7D%7D%7B%5Ctextrm%7Bliters%20of%20solution%7D%7D)
So, we have information about molarity (2M) and volume (80 ml=0.08 l), with that, we can find the moles of solute:
![\textrm{moles of solute}=\textrm{molarity}*\textrm{liters of solution}](https://tex.z-dn.net/?f=%5Ctextrm%7Bmoles%20of%20solute%7D%3D%5Ctextrm%7Bmolarity%7D%2A%5Ctextrm%7Bliters%20of%20solution%7D)
![\textrm{moles of solute}= 0.08 \textrm{ l} *2\textrm{ M} = 0.16 \textrm{ mol}](https://tex.z-dn.net/?f=%5Ctextrm%7Bmoles%20of%20solute%7D%3D%200.08%20%5Ctextrm%7B%20l%7D%20%2A2%5Ctextrm%7B%20M%7D%20%3D%200.16%20%5Ctextrm%7B%20mol%7D)
The mathematical equation that establishes the relationship between molar weight, mass and moles is:
![\textrm{molar weight}= \frac{\textrm{mass}}{\textrm{moles}}](https://tex.z-dn.net/?f=%5Ctextrm%7Bmolar%20weight%7D%3D%20%5Cfrac%7B%5Ctextrm%7Bmass%7D%7D%7B%5Ctextrm%7Bmoles%7D%7D)
![\textrm{MW}= \frac{\textrm{m}}{\textrm{n}}](https://tex.z-dn.net/?f=%5Ctextrm%7BMW%7D%3D%20%5Cfrac%7B%5Ctextrm%7Bm%7D%7D%7B%5Ctextrm%7Bn%7D%7D)
We have MW (58.44g/mole) and n (0.16 mol), and we need to find m (grams of salt needed) to solve the problem:
![\textrm{m} = \textrm{MW * n}= 58.44\frac{\textrm{g}}{\textrm{mol}} * 0.16 \textrm{ mol} = 9.35 \textrm{ g}](https://tex.z-dn.net/?f=%5Ctextrm%7Bm%7D%20%3D%20%5Ctextrm%7BMW%20%2A%20n%7D%3D%2058.44%5Cfrac%7B%5Ctextrm%7Bg%7D%7D%7B%5Ctextrm%7Bmol%7D%7D%20%2A%200.16%20%5Ctextrm%7B%20mol%7D%20%3D%209.35%20%5Ctextrm%7B%20g%7D)